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Given the non-homogenous higher order differential equation

[tex]u'' - 2 u' - 8 u = 6 - 8t[/tex]

with the intial conditions u(1) = 0 and u'(1) = 1.

Find the solution.

I'm told by my professor that I'm supposed to use the fact

[tex]{u''} + {a_1} u' + {a_0} u = f [/tex]

The characteristic polynomial for this differential equation is supposedly

[tex]z^2 + {a_1} z + {a_0} z = 0[/tex]

Using this fact I find root of the polynomial of the differential equation above to be [tex]\lambda _{1,2} = 4,-2[/tex]

The I'm suppose to use the fact that (Existence and Uniqueness theorem of Non-liniear differential equations) if [tex]\lambda {_1} \neq \lambda {_2}[/tex]

[tex]u_0 (t) : = \int \limit_{t_0} ^t \frac{e^{\lambda_1 (t-s)} - e^{\lambda_2 (t-s)}}{\lambda_1 - \lambda_2} f(s) ds[/tex]

[tex]t \in I[/tex]

My solution.

Since lambda 1 and lambda 2 er different then there exist a unique partiqular solution in the form:

[tex]u(t) = C_1 e^{6-8t} + C_2 e^{-8} = 0[/tex]

By inserting part 1 of the condition into the above, I get that C_1 = C_2 = 0.

initial condition 1 = true.

I differentiate u(t) and get [tex]u'(t) = -8 e^{-2} C_1 = 1[/tex]

here I get that [tex]C_1 = \frac{1}{-8{e^{-2}}} = \frac{-e^2}{8}[/tex] and this proves that u'(1) = 1

Am I on the right track here now?

Sincerley Yours

Fred

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# Differential equation question (HELP)

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